5 Power Series

Definition 5.1 A formal power series is an expression of the form

\[P=\sum_{n=0}^\iy a_nT^n \tag{5.1}\]

with coefficents $a_n\in\C.$ As no convergence is required, this is really just a sequence $(a_n)_{n\in\N}$ of complex numbers.

We call $a_0$ the constant term of $P.$ The order of $P$ is the smallest $n\in\N$ such that $a_n\neq0.$

Example 5.1

Every polynomial is a formal power series.

Definition 5.2 Let $\C \llbracket T\rrbracket$ be the set of all formal power series. Define the addition and multiplication of $P=\sum_{n=0}^\iy a_nT^n,$ $Q=\sum_{n=0}^\iy b_nT^n\in\C\llbracket T\rrbracket$ by \[P + Q=\sum_{n=0}^\iy(a_n+b_n)T^n, \tag{5.2}\] \[PQ=\sum_{n=0}^\iy c_nT^n, c_n=\sum_{i+j=n} a_ib_j. \tag{5.3}\]

Proposition 5.1

These operations make $\C\llbracket T\rrbracket$ a commutative ring with unit $1,$ the series with constant term $1$ and all higher coefficients zero.
$P\in\C\llbracket T\rrbracket$ has a multiplicative inverse $\iff$ the constant term of $P$ is non-zero.

Proof.

is straightforward and left as an exercise.
Having an inverse amounts to the existence of $Q=\sum_{n=0}^\iy b_nT^n$ such that $PQ=1.$ By Equation 5.3 this means \[a_0b_0=1 \qquad \sum_{i+j=n} a_ib_j=0 (\forall n>0). \tag{5.4}\] The first equations shows that $a_0$ is invertible, proving ‘$\implies$’. For ‘$\Longleftarrow$’, the idea is to use Equation 5.4 to construct the coefficients $b_n$ of the inverse inductively. Of course, $b_0=a_0^{-1}.$ For $n=1,$ Equation 5.4 gives $a_0b_1+a_1b_0=0$ so \[b_1=-a_0^{-2}a_1.\]

For $n=2,$ $a_0b_2+a_1b_1+a_2b_0=0.$ Inserting the known $b_0, b_1$ implies \[b_2=a_0^{-3}(a_1^2-a_0a_2).\]

For general $n,$ given $b_0,\ldots, b_{n-1}$ we rearrange Equation 5.4 as \[b_n = -a_0^{-1}\sum_{j=0}^{n-1}a_{n-j}b_j,\]

which defines $b_n$ by recursion.

It is unnecessary to remember Equation 5.4. It is enough to recall the ansatz $PQ=1$ and the strategy of computing the coefficients of $Q$ inductively.

Example 5.2

Let $P=1-T.$ Then $P^{-1}=\sum_{n=0}^\iy T^n$ is the geometric series. We verify $(1-T)\sum_{n=0}^\iy T^n=\sum_{n=0}^\iy T^n-\sum_{n=1}^\iy T^n=1$ (telescope sum).

Definition 5.3 The domain $D(P)$ of a formal power series Equation 5.1 is the set of all $z\in\C$ such that the series of complex numbers \[P(z)=\sum_{n=0}^\iy a_nz^n,\]

obtained by substituting $T$ by $z,$ converges. We obtain a complex function \[D(P)\longra\C, z\longmapsto P(z).\]

More generally, fix a center $z_0\in\C.$ We then have a complex function

\[D(P,z_0)\coloneqq z_0+D(P)\longra\C, z\longmapsto P(z-z_0)=\sum_{n=0}^\iy a_n(z-z_0)^n \tag{5.5}\]

which differs from $P(z)$ only by a translation.

Example 5.3

The geometric series $P=\sum_{n=0}^\iy T^n$ converges for all $|z|<1$ since the partial sums $P_n(z)=\frac{z^{n+1}-1}{z-1}$ tend to $\frac{1}{1-z}$ as $n\to\iy.$ The geometric series diverges for all $|z|\geqslant1,$ since in this case $(z^n)_{n\in\N}$ is not a null sequence. Hence $D(P)=D_1(0).$

Recall the following concept from analysis.

Definition 5.4 For a complex function $f\colon D\to\C,$ the uniform norm is

\[\|f\|_{\iy,D}=\sup_{z\in D} |f(z)|. \tag{5.6}\]

A sequence of functions $(f_n)_{n\in\N}$ is uniformly convergent to $f$ on $D$ if \[\|f-f_n\|_{\iy,D}\to0 \text{ as } n\to\iy.\]

A series $\sum_{n=0}^\iy f_n(z)$ is a uniformly convergent series on $D$ if the sequence of partial sums $P_n(z)=\sum_{k=0}^nf_k(z)$ converges uniformly.

Uniform convergence on a subset $C\subset D$ refers to the uniform convergence of the functions restricted to $C.$

Theorem 5.1 Every formal power series $P$ has a unique radius of convergence $0\leqslant\rho\leqslant+\iy$ such that for every center $z_0$

\[D_\rho(z_0)\subset D(P,z_0)\subset\ol{D}_\rho(z_0). \tag{5.7}\]

In fact,

\[\rho=\sup\left\{r\geqslant0\ \middle|\ (|a_k|r^k)_{k\in\N}\text{ is a bounded sequence}\right\}. \tag{5.8}\]

Moreover, $P(z-z_0)$ converges absolutely and uniformly on every smaller disk $D_r(z_0)$ with $0<r<\rho$ and diverges at every point $z\in\C$ with $|z-z_0|>\rho.$

Proof.

For uniqueness, suppose $D_{\rho_\ell}(z_0)\subset D(P,z_0)\subset\ol{D}_{\rho_\ell}(z_0)$ for $\ell=1,2$ and $\rho_1\neq\rho_2,$ say $0\leqslant \rho_1<\rho_2.$ Pick $z\in\C$ with $\rho_1<|z-z_0|<\rho_2.$ Then $z\in D_{\rho_2}(z_0)\subset D(P,z_0)\subset\ol{D}_{\rho_1}(z_0),$ so $|z-z_0|\leqslant \rho_1,$ a contradiction.

It suffices to prove the last statement in the theorem for $\rho$ defined by Equation 5.8, because this implies Equation 5.7. For $z\in\C$ with $|z-z_0|>\rho$ the terms $a_k(z-z_0)^k$ do not tend to zero (they are unbounded), so the series $P(z-z_0)$ diverges.

Let $0<r<\rho.$ Using that $\rho$ is the least upper bound, there exists $r<s<\rho$ with $|a_k|s^k<B$ bounded. Pick $w\in\C$ with $s=|w-z_0|.$ For all $z\in D_r(z_0)$ we then have

\[\sum_{k=0}^\iy |a_k(z-z_0)^k|=\sum_{k=0}^\iy\Bigl|a_k(w-z_0)^k\Bigl(\underset{\mathclap{\smash{q=}}}{\underbrace{\frac{z-z_0}{w-z_0}}}\Bigr)^k\Bigr|\leqslant B\sum_{k=0}^\iy q^k. \tag{5.9}\]

Since $|q|<1,$ the geometric series on the right converges. Hence the comparison test implies that $P(z-z_0)$ converges absolutely. As for uniform convergence, let $P_n$ be the $n$-th partial sum. As in Equation 5.9, for all $z\in D_r(z_0)$ \[|P(z-z_0)-P_n(z-z_0)|=\left|\sum_{k=n}^\iy a_k(z-z_0)^k\right|\leqslant B\sum_{k=n}^\iy q^k.\]

The right hand side is independent of $z,$ so $\|P-P_n\|_{\iy,D_r(z_0)}\xrightarrow{n\to\iy}0.$

Remark 5.2. The domain $D(P)$ of a formal power series is roughly a disk of radius $\rho,$ with uncertain behavior on the boundary circle $|z|=\rho.$ Determining the behavior on the boundary can be very difficult. On the other hand, the radius of convergence is easy to compute. If you find $z\in\C$ for which $P(z)$ converges (for example, using the ratio or the root test), you can deduce $|z|<\rho.$ If $P(z)$ diverges at $z\in\C,$ you know $|z|\geqslant\rho.$

Example 5.4 (optional)

For $P=\sum_{n=1}^\iy\frac{z^n}{n}$ we have $\rho=1$ by Equation 5.8. We investigate the behavior on the boundary circle $|z|=1.$ For $z=1,$ this is the divergent harmonic series. We claim that $P$ converges for all other boundary points, so $D(P)=\ol{D}_1(0)\setminus\{1\}.$ Consider \[\begin{align*} (z-1)\sum_{k=1}^n\frac{z^k}{k}&=\sum_{k=2}^{n+1}\frac{z^k}{k-1}-\sum_{k=1}^n\frac{z^k}{k}\\ &=\frac{z^{n+1}}{n} - z + \sum_{k=2}^n\left(\frac{z^k}{k-1}-\frac{z^k}{k}\right)\\ &=\frac{z^{n+1}}{n} - z + \sum_{k=2}^n\frac{z^k}{k(k-1)}. \end{align*}\] The series $\sum_{k=2}^\iy\frac{z^k}{k(k-1)}$ is (absolutely) convergent for $|z|\leqslant1.$ It follows that the left hand sequence of partial sums converges as $n\to\iy.$ Dividing by $z-1\neq0$ implies that $P$ converges.

Corollary 5.1 The restriction of the complex function $f(z)=P(z-z_0)$ defined in Equation 5.5 to $D_\rho(z_0)\subset D(P,z_0)$ is continuous.

Proof.

Since the translation $z\mapsto z-z_0$ is a homeomorphism, it suffices to consider the case $z_0=0.$ We prove continuity at each $w\in D_\rho(0).$ Let $\ep>0.$ By uniform convergence, we can pick $n\in\N$ with $\|P_n-P\|_{\iy,D_\rho(0)}<\ep/3.$ Since the polynomial $P_n(z)$ is continuous, there is $\de>0$ such that $|P_n(z)-P_n(w)|<\ep/3$ for all $z\in D_\de(w).$ By shrinking $\de,$ we may suppose $D_\de(w)\subset D_\rho(0).$ Then have \[|P(z)-P(w)|\leqslant|P(z)-P_n(z)|+|P_n(z)-P_n(w)|+|P_n(w)-P(w)|<\ep.\]

for all $z\in D_\de(w).$

Remark 5.2. Continuity need not hold on $D(P,z_0)$ (Sierpinski, 1916).

Example 5.5

The series $P=\sum_{n=1}^\iy\frac{z^n}{n^2}$ has $\rho=1.$ On the boundary circle, $|z^n/n^2|\leqslant1/n^2$ and the series $\sum_{n=1}^\iy\frac{1}{n^2}$ converges and is independent of $z.$ This implies uniform convergence and the proof of Corollary 5.1 shows that $P$ is continuous on its domain $D(P)=\ol{D}_1(0).$

When the radius of convergence is positive, we drop the word ‘formal’ and simply speak of a (convergent) power series.

Example 5.6

The exponential series \[\exp(z)=\sum_{n=0}^\iy \frac{z^n}{n!}\]

has radius of convergence $+\iy,$ since $\frac{z^n}{n!}$ is bounded (even tends to zero), as factorials grow faster than powers. Alternatively, for each $z\in\C$ we have absolute convergence by the ratio test \[\left|\frac{z^{n+1}/(n+1)!}{z^n/n!}\right|=\frac{|z|}{n+1}\longra0 \text{as} n\to\iy.\]

Therefore the power series defines a complex function $\exp\colon\C\to\C.$ Mathematically, this is a good definition for the exponential function. Using Equation 2.3 we obtain from this the series expansions of the sine and the cosine:

\[\cos(z)=\sum_{n=0}^\iy(-1)^n\frac{z^{2n}}{(2n)!}, \qquad \sin(z)=\sum_{n=0}^\iy(-1)^n\frac{z^{2n+1}}{(2n+1)!}.\]

Example 5.7 (optional)

We first generalize the definition of binomial coefficients to complex numbers $\al\neq0$ and $k\in\N.$ Set\[\binom{\al}{k}= \begin{cases} \frac{\al(\al-1)\cdots(\al-k+1)}{k!} &\text{if }k>0,\\ 1 &\text{if } k=0. \end{cases}\]The binomial series is \[B_\al = \sum_{k=0}^\iy \binom{\al}{k}z^k.\]

For $\al\in\C\setminus\N$ we have \[\left|\frac{\binom{\al}{k+1}z^{k+1}}{\binom{\al}{k}z^k}\right|=\left|z\frac{\al-k}{k+1}\right|\longra|z| \text{ as } k\to\iy.\]

Therefore, the ratio test implies $\rho=1$ when $\al\notin\N.$ When $\al\in\N$ only finitely many binomial coefficients $\binom{\al}{k}$ are non-zero, so $B_\al$ is a polynomial and $\rho=+\iy$ and indeed by the binomial theorem we have \[B_\al(z)=(z+1)^\al,\qquad\forall\al\in\N.\]

Theorem 5.2 Let $P$ be a formal power series. Assume the radius of convergence $\rho>0$ is positive. Then for each center $z_0\in\C$ the function \[P\colon D_\rho(z_0)\longra\C, z\longmapsto P(z)=\sum_{n=0}^\iy a_n(z-z_0)^n \tag{5.10}\]

is holomorphic with derivative given by termwise differentiation,

\[P'(z)=\sum_{n=1}^\iy na_n(z-z_0)^{n-1}. \tag{5.11}\]

Hence $P$ is infinitely complex differentiable, by induction.

Proof.

We show that $P$ is complex differentiable at every $w\in D_\rho(z_0).$ As $\sum_{k=0}^\iy kq^k$ converges for $|q|<1,$ the same argument given in Equation 5.9 shows that Equation 5.11 is a convergent series of complex numbers. For simplicity of notation, suppose $z_0=0.$ We have absolutely convergent series \[\begin{align*} \frac{P(w+h)-P(w)}{h}-P'(w)&=\sum_{n=0}^\iy a_n\frac{(w+h)^n-w^n}{h}-P'(w)\\ &=\sum_{n=0}^\iy a_n\left(\sum_{k=1}^n \binom{n}{k} w^{n-k}h^{k-1}\right)-P'(w)\\ &=\sum_{n=0}^\iy\sum_{k=2}^n a_n\binom{n}{k} w^{n-k}h^{k-1}\\ &=\sum_{k=2}^\iy\sum_{n=0}^\iy a_n\binom{n}{k} w^{n-k}h^{k-1}, \end{align*}\] where we use $\binom{n}{k}=0$ for $k>n$ and where we have exchanged the order of summation, by absolute convergence. The above is a power series $R(z)=\sum_{k=0}^\iy c_kz^k$ with $c_{k-1}=\sum_{n=0}^\iy a_n\binom{n}{k} w^{n-k}$ for $k\geqslant 2$ and $c_0=0.$ Our calculation shows that $R(h)$ converges, so the radius of convergence of $R$ is at least $h.$ Therefore $R(h)$ is continuous on $D_h(z_0),$ so $\lim_{h\to0} R(h)=R(0)=c_0=0.$ In other words, $\frac{P(w+h)-P(w)}{h}-P'(w)=R(h)\longra 0$ as $h\to0.$

The identity theorem for power series is the following result.

Corollary 5.2 Let $P=\sum_{n=0}^\iy a_n z^n,$ $Q=\sum_{n=0}^\iy b_n z^n$ be formal power series with positive radius of convergence. Let $0\neq z_\ell\in D(P)\cap D(Q)$ be a null sequence, $\lim_{\ell\to\iy} z_\ell=0,$ of non-zero complex numbers in the common domain. If $P(z_\ell)=Q(z_\ell)$ for all $\ell,$ then all coefficients $a_n=b_n$ agree.

Proof.

We prove this by induction.

Base case $n=0.$ By continuity of $P$ and $Q$ at $0,$ \[\begin{align*} a_0&=P(0)=P\left(\lim_{\ell\to\iy} z_\ell\right)=\lim_{\ell\to\iy} P(z_\ell)\\ &=\lim_{\ell\to\iy} Q(z_\ell)=Q\left(\lim_{\ell\to\iy} z_\ell\right)=Q(0)=b_0. \end{align*}\] Inductive step $n+1.$ Assume that $a_0=b_0, \ldots, a_n=b_n.$ By subtracting $\sum_{k=0}^n a_kz_\ell^k=\sum_{k=0}^n b_kz_\ell^k$ from $P(z_\ell)$ and $Q(z_\ell)$ we find \[\sum_{k=n+1}^\iy a_kz_\ell^k = \sum_{k=n+1}^\iy b_kz_\ell^k.\]

As $z_\ell\neq 0,$ we can divide this equation by $z_\ell^{n+1}$ and get $\tilde{P}(z_\ell)=\tilde{Q}(z_\ell)$ for the formal power series $\tilde{P}=\sum_{k=0}^\iy a_{k+n+1}z_\ell^k$ and $\tilde{Q}=\sum_{k=0}^\iy b_{k+n+1}z_\ell^k,$ which have the same (positive) radius of convergence. An application of the base case to $\tilde{P}$ and $\tilde{Q}$ then shows that $a_{n+1}=b_{n+1}$ for the constant terms, as required.

Questions for further discussion

Give precise statements of the comparison test, the ratio test and root test. Recall how the ratio test is proven by comparison with the geometric series.
Find a power series $P$ with $D(P)=\ol{D}_1(0)\setminus\{\pm1,\pm i\}.$
Explain why ‘is a bounded sequence’ in Equation 5.8 can be replaced by ‘is a null sequence’.

5.1 Exercises

Exercise 5.1

Find the radius of convergence for the following power series centered at the origin.

$\sum_{n=0}^\iy\frac{z^n}{n^3}$,
$\sum_{n=0}^\iy z^{3n}$,
$\sum_{n=0}^\iy\frac{z^n}{n^n}.$

Hint: Recall the ratio and root tests for series of complex numbers

Exercise 5.2

Treating $e^z,$ $\sin(z),$ $\cos(z)$ as formal power series with their usual Taylor expansion, find the terms of order $\leq3$ of the following power series:

$e^z\sin(z)$,
$\sin(z)\cos(z)$,
$1/\cos(z)$

Exercise 5.3

Let $P=\sum_{n=0}^\iy a_n z^n,$ $Q=\sum_{n=0}^\iy b_n z^n$ be power series with positive radii of convergence $\rho_P, \rho_Q>0.$ Show that:

$P+Q=\sum_{n=0}^\iy (a_n+b_n) z^n$ has radius of convergence $\rho\geqslant\min(\rho_P,\rho_Q).$
$PQ=\sum_{n=0}^\iy\left(\sum_{i+j=n}a_ib_j\right)z^n$ has radius of convergence $\rho\geqslant\min(\rho_P,\rho_Q).$

Exercise 5.4

Find a solution to the non-linear differential equation \[f'(z)+f(z)^2=0,\qquad f(0)=1\] on a disk centered at $z_0=0$ by making the ansatz $f(z)=\sum_{n=0}^\iy a_nz^n,$ inductively determining the coefficients $a_n,$ and finding the radius of convergence.

Exercise 5.5

Let $P(z)=\sum_{n=0}^\iy a_n z^n$ be a power series centered at $z_0=0$ and assume that the radius of convergence $\rho>0$ is positive. Suppose that $P(z)\in\R$ for all $z\in D_\rho(0)\cap\R.$ Prove that all coefficients $a_n,$ $n\in\N,$ must be real numbers. Deduce that $\ol{P(z)}=P(\ol{z})$ for all $z\in D_\rho(0).$

Exercise 5.6

Prove that the ring of formal power series $\C\llbracket T\rrbracket$ is an integral domain. In other words, show that $PQ=0\implies P=0$ or $Q=0.$

Exercise 5.7

The binomial coefficient of a complex number $\al\neq0$ and $k\in\N$ is defined as \[\binom{\al}{k}= \begin{cases} \frac{\al(\al-1)\cdots(\al-k+1)}{k!} &\text{if }k>0,\\ 1 &\text{if } k=0. \end{cases}\] The is the formal power series \[B_\al = \sum_{k=0}^\iy \binom{\al}{k}T^k.\]

Determine the radius of convergence of $B_\al.$
Prove that if $\al\in\N,$ then $B_\al(z)=(z+1)^\al.$
Show the generalized Vandermonde identity for $\al,\be,\al+\be\in\C^\t,$ \[\sum_{k=0}^n\binom{\al}{k}\binom{\be}{n-k}=\binom{\al+\be}{n}.\tag{$\star$}\]
Prove that for $\al=1/k$ the complex function $B_\al(z)$ satisfies $B_\al(z)^k=z+1$ on its domain. Hence $B_\al(z)$ is a $k$-th root of the function $z+1.$

Exercise 5.8

For $a>0$ and $z\in\C$ define $a^z=\exp(\log(a)z).$ Show that:

$a^zb^z=(ab)^z$ for all $a,b>0,$ $z\in\C$
$a^za^w=a^{z+w}$ for all $a>0,$ $z,w\in\C$
$|a^z|=a^{\Re(z)}$ for all $a>0,$ $z\in\C$

The is defined as \[\ze(z)=\sum_{n=1}^\iy\frac{1}{n^z}\qquad\text{for $\Re(z)>1.$}\tag{$\star$}\]

Prove that the series ($\star$) converges absolutely for all $z\in\C$ with $\Re(z)>1$ and uniformly on every subset $S_\de=\{z\in\C\mid\Re(z)>1+\de\}$ with $\de>0.$